If A+B+C=2S,then sin(S-A) cos(S-B) - sin(S-C) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $A+B+C=2S$,so $S-C = S-(2S-A-B) = A+B-S$. Also $2S-C = A+B$.The expression is $E = \sin(S-A) \cos(S-B) - \sin(S-C) \cos S$.Multiply and divi

Vedclass · Accepted Answer

$\cos A \sin B$

Vedclass · Answer

(C) Given $A+B+C=2S$,so $S-C = S-(2S-A-B) = A+B-S$. Also $2S-C = A+B$.The expression is $E = \sin(S-A) \cos(S-B) - \sin(S-C) \cos S$.Multiply and divide by $2$: $E = \frac{1}{2} [2 \sin(S-A) \cos(S-B) - 2 \sin(S-C) \cos S]$.Using $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$,we get:$E = \frac{1}{2} [(\sin(2S-A-B) + \sin(B-A)) - (\sin(2S-C) + \sin(S-C-S))]$.Since $2S-A-B = C$ and $2S-C = A+B$,we have:$E = \frac{1}{2} [\sin C + \sin(B-A) - \sin(A+B) + \sin C]$.This expression simplifies to $\sin B \cos A$.

If $A+B+C=2S$,then $\sin(S-A) \cos(S-B) - \sin(S-C) \cos S =$

Similar Questions

If $f(x) = \sin^6 x + \cos^6 x$ for $x \in R$,then $f(x)$ lies in the interval

The number of integral values of $k$ for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has a solution is:

If $M_1$ and $M_2$ are the maximum values of $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$ and $3 \cos^2 5x + 4 \sin^2 5x$ respectively,then $\frac{M_1}{M_2} = $

The value of $k$,for which $(\cos x + \sin x)^2 + k \sin x \cos x - 1 = 0$ is an identity,is

The maximum value of the function $f(x) = 3 \sin x + 4 \cos x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module